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If $\sum x_n$ converges, is the sequence $a_k =\sum_{n=1}^k x_{2n}^2$ Cauchy?

I suspect that it isn't but a bit stuck on finding a counter example. For a valid counterexample, I believe I need a series that converges but the squared even partial sum does not.

Writing the definition of Cauchy, we know a sequence $(x_n)$ is Cauchy if for every $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that $m, n \geq N \implies |x_m -x_n| < \epsilon$. But it seems easier to just use that convergence implies Cauchy rather than work directly from the definition. Any ideas/tips on what exactly I'm not seeing?

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  • $\begingroup$ You cannot find a counter-example if the sequence is nonnegative since then $x_n^2 \leq x_n$ for all sufficiently large $n$. So... $\endgroup$ – Michael Oct 15 '18 at 1:35
  • $\begingroup$ If $x_n=(-1)^n/\sqrt n$ then $\sum_n x_n$ converges but $\sum_n x_n^2$ diverges. (Yes, I am aware that this isn't your exact problem.) $\endgroup$ – Lord Shark the Unknown Oct 15 '18 at 1:35
  • $\begingroup$ @Michael What if $x_n<0$? $\endgroup$ – Lord Shark the Unknown Oct 15 '18 at 1:37
  • $\begingroup$ @LordSharktheUnknown : I don't follow your comment directed to me ("@Michael what if..."), I think you meant something else or you misread my first comment. Your first comment seems in teh same spirit as mine but gives a bit of a larger hint. $\endgroup$ – Michael Oct 15 '18 at 1:38
  • $\begingroup$ I see now there is a third hinter. With all these "hints" the problem will soon be solved without the asker's involvement! $\endgroup$ – Michael Oct 15 '18 at 1:45
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Hint: Let $\{y_n\}$ be a sequence of positive reals that converges to $0$, and take

$$x_n=\begin{cases} -y_{n/2} & \mathrm{if\ }n\equiv 0\bmod 2 \\ y_{(n+1)/2} & \mathrm{if\ }n\equiv 1\bmod 2,\end{cases}$$

or in other words $x_1,x_2,\cdots,$ is $y_1,-y_1,y_2,-y_2,y_3,-y_3\cdots$.

Then, your second series is

$$\sum_{n=1}^k y_n^2.$$

Can you find a sequence $\{y_n\}$ that converges to $0$ for which the above does not converge?

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  • $\begingroup$ This doesn't converge to 0, but it seems this would work for the question, $\sum \frac{(-1)^{n+1}}{n}$? Converges but the above squared doesn't converge. $\endgroup$ – SS' Oct 15 '18 at 2:12
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    $\begingroup$ @SS' Isn't the square of it $\sum \frac{1}{n^2}$, which converges? However, that idea almost works - what if you put a slower-growing function in the denominator? $\endgroup$ – Carl Schildkraut Oct 15 '18 at 2:46

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